c08，c09二维数组，自己的字符串函数，strcpy，strcat，strncat，strncpy，atoi，itoa，strcmp

二维数组，魔方阵，杨辉三角

1.打印函数

void Show(int (*brr)[3],int row,int col)
{
	for(int i = 0;i < row;i++)
	{
		for(int j = 0;j < col;j++)
		{
			printf("%d ",brr[i][j]);
		}
		printf("\n");
	}
}

void Show2(int *brr,int row,int col)//方法二
{
	for(int i = 0;i < row;i++)//1
	{
		for(int j = 0;j < col;j++)//3
		{
			printf("%2d ",brr[i*col+j]);//1*3+0
		}
		printf("\n");
	}
}

int main()
{
	int brr[ROW][ROW] = {};
	Fun2(brr);
	Show2((int *)brr,ROW,ROW);
}

2.对称矩阵

//对称矩阵
void Fun2(int (*brr)[ROW])
{
	srand(time(0));//随机种子
	for(int i = 0;i < ROW;i++)
	{
		for(int j = 0;j < ROW;j++)
		{
			if(j > i)
			{
				brr[i][j] = rand()%10+1;//[0-n)
			}
			else if(i == j)
			{
				brr[i][j] = 0;
			}
			else
			{
				brr[i][j] = brr[j][i] ;
			}
		}
	}
}

注意
//int (*brr)[3] 数组指针
//int *brr[3] 指针数组

3.注意

int main()
 { 
	int brr[2][3] = {1,2,3,4,5,6};
	printf("brr == %d\n",brr);
	printf("brr+1 ==%d\n",brr+1);
	printf("*brr+1 == %d\n",*brr+1);
	printf("*(brr+1) == %d\n",*(brr+1));
	printf("*(brr+1)+2===>%d\n",*(brr+1)+2);
	printf("&brr[1][2]==>%d\n",&brr[1][2]);
	printf("*(*(brr+1)+2)====%d\n",*(*(brr+1)+2));
	printf("*(brr[1]+2)===>%d\n",*(brr[1]+2));  
	return 0；
	}

4. 练习题（杨辉三角，魔方阵）

#include<stdio.h>
#include<assert.h>
#include<assert.h>
#define ROW 3
void Show2(int *brr,int row,int col)
{
	for(int i=0;i<row;i++)
	{
		for(int j=0;j<=i;j++)
		{
			printf("%2d",brr[i*row+j]);
		}
		printf("\n");
	}
}
void Show(int *brr,int row,int col)
{
	for (int i = 0; i < ROW; i++)     
	{
         for (int j = 0; j < ROW; j++)
        {
			printf("%3d",brr[i*row+j]);
         }
        printf("\n");   //打印一行回车 
	}
}
//作业杨辉三角
void YangHuiTriangle(int (*brr)[ROW])//数组指针
{
	brr[0][0]=1;
	for(int i=1;i<ROW;i++)//i=0;
	{
		for(int j=0;j<ROW;j++)//j<=i;	
		{
			if (j==0)//第0列为1
			{
				brr[i][j]=1;
			}
			else if(i==j)//对角线为1
			{
				brr[i][j]=1;//brr=brr+i;
			}
			else 
			{
				brr[i][j]=brr[i-1][j-1]+brr[i-1][j];//杨辉三角的公式
			}
		}
	}
}

魔方阵
  //魔方阵奇数行列对角线加起来相等，1放在第0行中间位置。每次放在上一行下一列，如果当前数字的上一行下一列列有数据，则放在数字的下一行
    void MagicMquare1(int (*brr)[ROW])
    {
    	assert(ROW%2!=0);
    	int count =1;
    	int row=0;
    	int col=ROW/2;
    	while(count<=ROW*ROW)
    	{
    		brr[row][col]=count;
    		 int i=row;
    		 int j=col;
    		if(i==0)//如果i为0，则放到ROW-1行
    		{
    			i=ROW-1;
    		}
    		else//i!=0，放到i-1行
    		{
    			i=i-1;
    			//j=j+1;
    		}
    		j=(j+1)%ROW;//保证j是有效行,即j为可以+1的行
    		if(brr[i][j]!=0||(row==0&&col==ROW-1))//如果有值，放在当前的下一行
    		{
    			i=row+1;
    			j=col;	
    		}
    		row=i;
    		col=j;
    		count ++;
    	}
    	/*for (int i = 0; i < ROW; i++)     
    	{
             for (int j = 0; j < ROW; j++)
            {
    			printf("%3d",brr[i][j]);
             }
            printf("\n");    
    	}*/
    }

    int main()
    {
    	int brr[ROW][ROW]={0};
    	//YangHuiTriangle(brr);
    	//Show2((int *)brr,5,5);
    	MagicMquare1(brr);
    	Show((int *)brr,3,3);
    }

void MagicSqaure(int (*arr)[ROW])//法二
{
	assert(ROW % 2 != 0);
	arr[0][ROW/2] = 1;

	int preRow = 0;
	int preCol = ROW/2;

	for(int i = 2;i <= ROW*ROW;i++)
	{
		if( arr[(preRow-1+ROW) %ROW] [(preCol+1)%ROW] != 0 )
		{
			preRow = (preRow+1) % ROW ;
		}
		else
		{
			preRow = (preRow-1+ROW) %ROW;
			preCol = (preCol+1)%ROW;
		}
		arr[preRow][preCol] = i;
	}
}

5.My_strcpy，strcat，strncat，strncpy，atoi，itoa，strcmp

5.1My_strcpy，My strncpy

char * My_strcpy4(char *dest,const char *src)
{
	//char *p = dest;
	assert(dest != NULL && src != NULL);
	while(*dest++ = *src++) {}
	return dest;//走到最后\0
	//return p;
}
#include<stdio.h>
#include<String.h>
#include<assert.h>
char *My_strncpy( char *dest,char *src,int n)
{
	char *p=dest;
	int count=0;
	assert(dest!=NULL&&src!=NULL);
	assert(n>0&&n<strlen(src));
	 while(count<n)
	 {
		 *dest++=*src++;
		 count++;
	 }
	 return p;
}
int main()
{
	char dest[10]="";
	printf("%s\n",My_strncpy(dest,"abcdefgh",5));
	return 0;
}

5.2My strcat，My strncat

char *My_strcat(char *dest,char *src,int len)
{
	assert(dest!=NULL&&src!=NULL);
	if(strlen(dest)+strlen(src)>len);
	char *p=dest;
	while(*dest!='\0')
	{
		dest++;	 
	}
	while(*dest++=*src++);
	return p;
}
int main()
{
	char dest[10]="abc";
	char *src="def";
	int len=sizeof(dest)/sizeof(dest)[0];
	printf("%s\n",My_strcat(dest,src,len));
	return 0;
}

5.3My strcmp

int Mystrcmp(const char *str1,const char *str2)
{
	assert(str1!=NULL&&str2!=NULL);
	int tmp=0;
	while((tmp=*str1-*str2)==0&&*str2!='\0')
	{
		str1++;
		str2++;
	}
	return tmp;
}
int main()
{
	char *str1="adefs";
	char *str2="aadek";
	printf("%d\n",Mystrcmp(str1,str2));
	return 0;
}

5.4 atoi，itoa

#include<stdio.h>
#include<string.h>
#include<assert.h>
#include<ctype.h>
 
int My_atoi(char *str)
{
	int p=0;
	assert(str!=NULL);
	while(*str==' '||!isdigit(*str))
	{
		str++;
	}
	while(*str=='-')
	{
		printf("%s\n",'-');
	}
	while(isdigit(*str))
	{
		p=p*10+(*str-'0');
		str++;
	}
	return p;
}
int main()
{
	printf("%d\n",My_atoi(" daa-123"));
	return 0;
}
char *My_itoa(char *str,int num) 
{
	int i=0; 
	if(num<0)
	{
		num=-num;
	}
	while(num!=0)
	{
		str[i]=num%10+'0';
		i++;
		num/=10;
	}
	str[i]='\0';
	i--;
	for(int j=0;j<i;j++,i--)
	{
		char ch=str[j];
		str[j]=str[i];
		str[i]=ch;
	}
	return str;
}
int main()
{
	char str[10]=" ";
	printf("%s\n",My_itoa(str,123));
	return 0;
}

5.5找出和为给定数值的两个元素的下标

void Sumk(int *arr,int key,int *num1,int *num2,int len)
{
	int low = 0;
	int high = len-1;
	while(low < high)
	{
		if(arr[low]+arr[high] > key)
		{
			high--;
		}
		else if(arr[low]+arr[high] < key)
		{
			low++;
		}
		else
		{
			*num1 = low;
			*num2 = high;
			break;
		}
	}
}
	 
int main()
{
	int arr[]={4,6,8,9,10,11,13,23};
	int len=sizeof(arr)/sizeof(arr[0]);
	int index1 = 0;
	int index2 = 0;
	Sumk(arr,20,&index1,&index2,len);
	printf("%d,%d\n",index1,index2);
	//Print();
	//printf("%d\n",CheckSum());
	return 0;
}

5.6统计单词的个数

#include<stdio.h>
#include<assert.h>
#include<ctype.h>
int NumWord(char *str)
{
	 assert(str!=NULL);
	 int count=0;
	 while(*str!='\0')
	 {
		 if(isalpha(*str)&&!isalpha(*str+1))
		 {
			 count++;
		 }
		 str++;
	 }
}
int main()
{
	printf("%\c",NumWord("dgs jsb jkd sjd sk jd k"));
}